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3 years ago

The capacitor in the figure (Figure 1) is initially uncharged. The switch is closed at t=0.

Physics

1 answer:

fomenos3 years ago

3 0

Immediately after the switch is closed, the capacitor acts like a short circuit for a very small time. We can calculate the currents through each resistor by first calculating the total resistance:

Req = 1/(1/3+1/6) + 8 = 1/(3/6) + 8 = 2 + 8 = 10 ohms.
IR1 = E / Req = 42 / 10 = 4.2 A.
VR2 = VR3 = E * R23 / Req = 42 * 2 / 10 = 8.4 V
IR2 = VR2 / R2 = 8.4 / 6 = 1.4 A.
IR3 = VR3 / R3 = 8.4 / 3 = 2.8 A.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

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The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

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